finite element modelling · introduction to fem: stiffness equations fora axial bar element in...

FINITE ELEMENT MODELLINGCourse code:AME014

VI- semesterRegulation: IARE R-16

Mrs V. PrasannaAssistant Professor

DEPARTMENT OF MECHANICAL ENGINEERINGINSTITUTE OF AERONAUTICAL ENGINEERING

(Autonomous)DUNDIGAL, HYDERABAD - 500 043

COS COURSE OUTCOMES

CO1 Describe the concept of FEM and difference between the FEM with other methods and problems based on 1-D bar elements and shape functions.

CO2 Derive elemental properties and shape functions for truss and beam elements and related problems.

CO3 Understand the concept deriving the elemental matrix and solving the basic problems of CST and axi-symmetric solids

CO4 Explore the concept of steady state heat transfer in fin and composite slab

CO5 Understand the concept of consistent and lumped mass models and solve the dynamic analysis of all types of elements.

UNIT-IINTRODUCTION TO FEM

Introduction to FEM:

Stiffness equations for a axial bar element in local co-ordinates using Potential Energy approach and Virtual energy principle.

Finite element analysis of uniform, stepped and tapered bars subjected to mechanical and thermal loads.

Assembly of Global stiffness matrix and load vector.

Quadratic shape functions.

Properties of stiffness matrix

Introduction

o DifferentialEquationd

EA(x) du

f (x) 0 0 x L

o Boundary Condition Types

Prescribed displacement (essential BC)

Prescribed force/derivative of displacement (natural BC)

Axially Loaded Bar – Governing Equations and Boundary Conditions

Examples

Fixed end

Simple support

Free end

Axially Loaded Bar –Boundary Conditions

Elastic Potential Energy (PE)

Spring case

Axially loaded bar

Unreformed:deformed:

Elastic bodyT

PE 1 σ εdv

Potential Energy

PE 2 Adx

Potential Energy

Work Potential (WE)

Total Potential Energy

f: distributed force over a line P: point forceu: displacement

2 Adx u fdx P uB

Principle of Minimum Potential Energy

For conservative systems, of all the kinematically admissibledisplacement fields, those corresponding to equilibrium extremizethe total potential energy. If the extremum condition is aminimum, the equilibrium state is stable.

Potential Energy + Rayleigh-Ritz Approach

Example:

Step 1: assume a displacement field

f is shape function / basis function

n is the order of approximation

Step 2: calculate total potentialenergy

u aii xi

i 1 to n

Potential Energy + Rayleigh-Ritz Approach

Example:

Step 3: select ai so that the total potential energy is minimum

Galerkin’s Method

Example:

d duEA( x) f (x) 0

Seek and approxdiu~mationso

dx dx i

w EA(x) dx dx

V f ( x)dV 0

EA(x)du

u~x 0 0

EA(x) Pdx xL

In the Galerkin’s method, the weight function is chosen to be the same as the shape function.

Galerkin’s Method

Example:

d du~ L

du~dwL

dx wiEA(x) dx

i dx i iw fdx w EA(x) du~ 0

f (x)dV 0

EA(x)dx dx

Differential Equationd

EA(x) du

f (x) 0

Weighted-Integral FormulationL

0 dx dx

w EA(x)

f (x)dx 0

Weak FormL dw du du L

0 dx EA(x)

dx wf (x)dx w EA(x)

FEM Formulation of Axially Loaded Bar – Governing Equations

Example:

Step 1: Discretization

Step 2: Weak form of one elementx2 dw du

dxEA(x)

dx w(x) f (x)dx w(x)EA(x)

x2 dw du dx EA(x)

dx w(x) f (x)dx w x2 P2 w x1 P1 0

Approximation Methods – Finite Element Method

Approximation Methods – Finite Element Metho

Example (cont):

Step 3: Choosing shape functions- linear shape functions

u 1u1 2u2

x x 1 1 1

2 ;l 2

2 x x 1; x l

Example (cont):

Step 4: Forming element equation

x2 1u u EA

Letw 1, weak form becomes

2 1f dx 1P21P1 01u 2 u f dx P1 1

weak form becomes x2 1

x2EA EA Let w ,

2 EA 2 1 dx 2 f dx 2P2 2P1 01 u

x1l l x1

u2 2 f dx P2

1 fdxEA 1 11 u

1P f P

2 P f P l 1 1 u

x1 fdx

Step 5: Assembling to form systemequation

1 I I I

0 u1 f1 P1

EI AI 1 uI

Approach 1:

Element 1: 2

l I 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0

E IIAII 0 0 uII f II PII

Element 2: 1 1

f II PII

l II 0 1 1 0 uII

0 0 0 0 0 0 0

E III AIII 0

0 0 0 0

Element 3:III 0

0 0 0 0

uIII f III PIII

l 1 1 1 1 0

0 1 1 uIII f III PIII

Example (cont):

lII lIII

EIIIAIII

f P4 4 4 2 2

Example (cont):

Step 5: Assembling to form system equation

Assembled System:

lI 0 0

E AI I I I II II u1 f1 P1 f1 P1 I

f II lI

E A EII AII

lI lII

lIIu2 f2 P2 f

2 1 P2 P1I II

EIIAII

EIIAII EIIIAIII

EIIIAIII

f II f III PII

0 3 3 3 2 f III

1 2 PIII1

Approximation Methods

Example (cont):

Step 5: Assembling to form systemequation

Approach 2: Element connectivity table

ke Kij IJ

Example (cont):

Step 6: Imposing boundary conditions and forming condense

E IIAII E II AII

Condensed system: EIAI

l I l II l II0

E IIAII

E III AIII

l IIE IIAII

E IIIAIII

l III l III u f 03 3

E IIIAIII

E III AIII 4 4

Step 7: solution

Step 8: post calculation

u u11 u22 u1

dx dx E Eu 1

dx Eu2

Example (cont):

Summary - Major Steps in FEM

Discretization

Derivation of element equation

• weak form

• construct form of approximation solution over oneelement

• derive finite element model

Assembling – putting elements together

Imposing boundary conditions

Solving equations

Post computation

Linear Formulation for Bar Element

K11 K12 u1P

2 2 12 22 u2

where K x2 EAdi j ddx K , f x2 f dx

ji i ix1

x=x1 xx=x2

Higher Order Formulation for Bar Element

u(x) u11(x) u22 (x) u33 (x)

u(x) u11( x ) u22 ( x ) u33( x ) u44 ( x )

u(x) u11( x ) u22( x ) u33( x ) u44( x ) unn( x )

Natural Coordinates and Interpolation Functions

4 16 3

27 1 1

1, 9 1 1 1

x x1 x2

Natural (or Normal) Coordinate: 2

2 1 2 3 1 ,

9 1 1 1,

27 1 1 11

Quadratic Formulation for Bar Element

12 22 23 2P f 2 2

K12 K13 u1K K

K u 13 23 33 3

3 3 x2 d d

1 d d 2

where Kij i j

d K ji EA

1 d d l

and fi i f dx i f 2d , i, j 1, 2, 3

Quadratic Formulation for Bar Element

1 1u 1u( ) u ( ) u (1 1 2 2

) u ( ) u3 3 1 2

2 3 11, 1

x x 1 x 2

2l / 2

d d xl

d1 1 ,d2

2 d 2 d3 3 dx l d

2 d 2 1

l dx l d

2 d 2 1

UNIT-IIANALYSIS OF TRUSSES AND BEAMS

Finite Element Analysis of Trusses:

Stiffness equations for a truss bar element oriented in 2D plane

Finite Element Analysis of Trusses

Plane Truss and Space Truss elements

Methods of assembly

INTRODUCTION

Arbitrarily Oriented 1-D Bar Element on 2-D Plane

sin0 0 u1 u1v 0

cos sin cos0 0 v

0 cos sin u2

sin cosv2

Relationship Between Local Coordinates and Global Coordinates

sin0 0 P1P1 cos0 sin cos0 0

1 0 co s sin

0 0 sin cosQ2

Stiffness Matrix of 1-D Bar Element on 2-D Plane

K AE 0 0

1 0 1 0

ij L 1 0 1 0

0 0 0 0

P1 cos2 sin cos cos2 sin cosu1 Q AE

sin cos sin2 2 sin cos sin v

cos sin cos u2 2

Q2 sin cos

sin cos

sin cos sin2 v2

P1 cos sin 0 0 cos sin 0 0 u1

Q1 sin

Arbitrarily Oriented 1-D Bar Element in 3-D Space

1 x x x

P ,u1 1

2 P2,u2

, , are the

Direction Cosines of the

bar in the x-y-z

coordinate system

P1 x 0

0 0 0 v 1 y

y y 11 y y y 1 R 0

x x 0 0

Q 0 0 0 0 Q

0 0w 0

1 z z z

0 0 0 w

1 P 0z 0z 0z P

u 0 0 0 u 2 x x x 22 x 2 Q

v2 0 y 2

0 y y v

Q 0 0 0 0

R2 0 0 0 0

z 22w 0

0 0 0 z z zw2

Stiffness Matrix of 1-D Bar Element in 3-D Space

z2 P2 ,u2

1 0 AE

y P2 L

P1 ,u1x

2 x x x

2 u1 x x x

x v1 x x2

x x 2 w1

AE x x x x x xx xx x P2

2 x x x x

2x x x x x

Q 2 2 v2

x x x x x x x x x R2 x x x x 2

x x x x2

01 0 0 0 0 0

0 0 0 1 0 00

0 0 0 0 00

0 0 0 0

AE3 3 3

P1 1 0 1

Q 0 0 0 v0u1

Element I 1 0u1 2 P2

L 1 0 1

1 Q AE 3 3 3 v

Element II 2 2

P3 4L 1 3u3

Element III

3 1 3u1

Matrix Assembly of Multiple Bar Elements

1 3 u23

P1 u1 Q

1 1 Element I

P AE u2

u1Q v 1

Element II2

P AE u2

Element III 1 12

4 0 4 0 0 0

0 0 0 0 0 0

4 0 4 0 0 0

0 0 0 0 0 0

0 0 1 3 1 3

0 0 3 3 3 3

0 0 1 3 1 3

0 0 3 3 3 3

3 3Q3 v3

R1 4 1 1 3 u 1

0 4 1 0 3 1 3

0 0 3 0 3 3 1 1

3 3 3 3 v3

Apply known boundary conditions

R1 ? u1 0

0 6S3 ? v3 0

Solution Procedures

S 0R2 F 5 3 1

3 v ?3 u1 0

0 1 3 u2 ?

S3 ? 3 3

1 3 1 3

6 v3 0

u1 0v 0

S1 0 1 4FL AE

4 0 5 1

3 4 3u2

R1 ? 4 L 3 3 5 AE

S3 ? 3 3

0 0 3 3

1 3 1 3

0 3 3 0 1

Recovery of axial forces

P1 1 0 u1 0 4

Element Q1 AE 0 1

L 1 0 1 0

Q2 0 0 0 0 v 0 0

u2Element Q2 AE 3 3 3 v 05AE

3u 0Q3

3 3 3 3 3v 0

P1 1 1 0Q

AE 3 3 3 3

3u1 0v 0

Element III

3 4L 1 3

Q3 3 3

3 3 v3 0

Stresses inside members

Element

Element III

Finite Element Analysis of Beams

Hermite shape functions

Element stiffness matrix

Load vector

Problems

Pure bending problems

Normal strain:x

Normal stress:

Normal stress with bending moment: x ydA M

Bending Beam

Moment-curvature relationship:

Flexure formula: x

M EI EI

I y2dA

1 d2 y

Review

Bending Beam

Review

Deflection

Relationship between shear force, bending moment and transverse load:

dV q dM V

EI q dx4

Sign convention:

Governing Equation and Boundary Condition

{Natural BCs – if v

dxv ? & ? & EI ?& EI

dx dx2 dx

2at x L

Essential BCs – if v or is specified at the boundary.

or is specified at the boundary.

d 2vEIdx2

d 2vEI

dx dx2

Governing Equationd 2

2d v(x)

q(x) 0,

Boundary Conditions

v ? &dv

? & EId 2v

?& at x 0

dx dx2 dx dx2

dv d 2v d d 2v ? ,dv

d EI d 2v ?,

Weak Form from Integration-by-Parts ----- (2nd time)

x2 d 2 w d 2v d x2

d 2v dwx2

d 2v 0 2

EI dx2 wqdxw EI dx2 dx EI dx2

x1 dx dx x x

x2 d 2w

x2 d 2 w d 2v dw x2

wqdx wV

EI dx2

Q1 Vx1, Q2 M x1, Q3 V x2, Q4 M x2

x2 d 2 w d 2v 2dx EI dx2 wqdx w(x )Q1 1

w(x2 )Q3

1x 1 2

Letj 1

v(x) u j j (x) and n 4

3 2 4 ;

4 x2 d 2w EIu

wqdx w(x )Q w(x )Q dw Q

dx2 j1

1 1 2 3 2 4

Let w(x)= fi (x),

d dx2 d

2iEI 4u

i = 1, 2, 3, 4

d 2j qdx (x )Q (x )Q

x1 dx 2 j1 j

i 1 1 i 2 3 dx 1 dx 2

Ritz Method for Approximation

u vx ; u dv

; u vx ;dx

dx 2x1Q

Qi x2 3

K u qij j i

where K x2 EI 2i d 2 x

dx2 dx2

dx and q 2 qdxj

Ritz Method for Approximation

2 2 Q1 K11 1

d1 x dx

x x2 dx

K14 u1 q dx

xQ2 u2 q2

x1 dx3

x2 dx3x Q K41 K42 K43

K34 u3 q3K44 u4 q4

x1 dx x

K12 K13

K21 K22 K23 K24

K32 K33

Selection of Shape Function

Q K 2 12 22 23

Q4 K14

13 23 33

K24 K34 K44 u4 q4

dx x x2 dx

dx x 1 x 2 3

Interpolation

Propertiesx1 dx x2 dx

0 1 0 0

0 0 1 00 0 0 1

1x x2 dx x

Q1 K11 K12 K13 K14 u1 q1

v ( ) u 11 u 2 2 u 3 3 u 4 4

and u1

ddv( )

where dv1

dv2u v u

1 1 2d

u v u3 2 4

Let i i a bi ci

Find coefficients to satisfy theinterpolation properties.

Derivation of Shape Function for Beam Element Local Coordinates

Derivation of Shape Function for Beam Element

In the global coordinates:l

dv1 (x) l

dv2 (x)v(x) v1 1

(x) v2 3

1 3 1 2 1

x x x2 x1 x2 x21 2

x x11 1

x x2 x1 x x

3 3 1 21

4 1 x2 x1 2 x x

x x 1 x

x x1x x

2 1 2 1

Element Equations of 4th Order 1-DModel

Q1 q1 K11 K12 K13 K14 u1Q q K u 2

2 12 22 23 24 2Q

K K K K 3 3

13 23 33 34

K14 K24 K34

where K x2 2 EI i x2 qdx

dx2 ji

K44 u4

Element Equations of 4th Order 1-DModel

Q1 q1 3L L2 u 2

Q q2 2 1

L3 3Lu v

6 3L 6 3L u1 v1 2L2 3L

6 3L 6

3L L2 3L2 2L u

Q3 4 4 4 2

where qi iqdxx1

Example 1.

Governing equation:

dx2 dx2

Weak form for one element

Finite Element Analysis of 1-D Problems - Applications

x1dx2 dx2

d w d v wq dx w x1Q1 dw

dxQ2 wx2Q3

dx 4Q 0x2

Q1 V(x 1) Q2 M (x1) Q3 V(x2) Q4 M (x2 )

d 2 EI d 2v q(x) 0

Finite Element Analysis of 1-D Problems

Approximation function: v(x) v (x) l dv1 (x) v (x) )

1 12 dx

2 2 32 dx

1 3 1 2 1

x12 x2 x2 x1

1x x 1x x1

l x2 x1

x x2 x x

3 3 1 2 1

4 x2 x 1 x2 x1 2 x x

l x x x x1 2 1 2x x1

Finite Element Analysis of 1-D Problems

Finite element model:

Q 2EI 3L L 2 2

L3 3Lv2

6 3L 6 3L v1

2L2 3L

6 3L 6

3L L2 3L 2L2 2

Discretization:

Matrix Assembly of Multiple Beam Elements

Q I 6 3L 6 3L 0 0 0 0 v

2 3L 2L 3L L 1Q I 2

Q I 2EI 3L

0 0 0 06 3L 6 3L 0 0 0 0 v

L2 3L 2L2 0 0 0 0 Element I

0 0 0 0 0 v30 0 0 0 0

0 0 0 0 0 v0 0 0 0 0

0 0 0 0 0 00

0 00 0 0 0 0 0 0 0

1II Q1

Element II 0 3L

3L 0 0 v2

2EI 0 0 0 2

L3 0 0 6

0 0 6 3L 6

2L2 3L

3LQ3II 6 3L 0 0 v3II

0 0 3L L2 3L 2L2 0 0 Q 4

30 0 v4 0

0 0 0 0 0 0

0 0 0 0 0 00 4

Matrix Assembly of Multiple Beam Elements

P1 6 3L 6 3L 0 0 0 0 v1 M 3L 2L 2 3L L2 0 1 0 0 0

P 2 6 3L 6 6 3L 3L 6 3L 0 0 v2 M

2EI 3L L2 3 L 3L 2L2 2L2 3L L2 0 0

2 2 P L3 0 0 6 3L 6 6 3 L 3L 6 3L

M 03 0 3L L2 3L 3L 2L2 2L2 3L L2

P 4 0 0 0 0 6 3L 6 3Lv4M

0 0 0 0 3L L2 3L 2L2

2EI 0 0

0 0 0 0 0 0 0 v1

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 v2 0 0 0 0 Element III

1 v36 3L2 6 3L2

QIII 3L

3 L 0 0 0 0

0 0 0 0 3L 2L 3L

0 0 0 6 63L

4Q III

0 0 0 0 3L L2 23L 2L 4

Apply known boundary conditionsP1 ? 6

M ? 3L3L 6 3L2L2

3L L2 0 0

P2 ? 2EI 3L L

0 4L12 0

2 6 3L 0

L2 3L 0 0

0 v2 0

L3 0 0 6 3L

M 0 2 2 3L L2 ?

P4 F 0

0 0 3L L

12 0 6 3L v 0

0M 4 0

3L L2 3L 2L2

3Lv4 ? ?

M 0M 2 0

3L L0 0 L2

L2 03 1

M 0 2EI 0 0 0 0 6

6 22 3Lv 0

M1 ? 2

0 4L 3L L 0 0 v1 03L 0 4L2 3L

0 0 3 ?

Solution Procedures

0 0 0 0 v1

0 0 0 01

L 6 3L 6

3L 2L 3L

6 3L 12 0 6 3L 0

0 0 6 3L 12 0 6

3L 4 ?

Solution Procedures

M 2 0 3L

L2 03 1

P4 F M 0 2EI 0

0 3L 4L2 L2 0 0 v1 03L 0 L2 4L2 3L

0 6 0 3L 60 3L 0 L2 3L 2L2 v

0 3Lv2 0

L 6 0 0 2

3L 6 02

P ? 3 4 ?

2M 0 0

? P ?? ? 2EI L2

P F L3 P ?

L3 v 3

4 4 2 4

6 3L v ?

3L 2L2

4 ? P ? 3

3L 0 0 00 0 0

0 3L 0 03L 0 6 3L 4

0 0 6 12 3L 0 6 3L

Shear Resultant & Bending Moment Diagram

UNIT-III2D ANALYSISENTS

Introduction

Computation of shape functions for constant straintriangle

Properties of the shape functions

Computation of strain-displacement matrix

Computation of element stiffness matrix

Computation of nodal loads due to body forces

Computation of nodal loads due to traction

Recommendations for use

Example problems

Finite element formulation for 2D

Divide the body into connected to each other through special points (“nodes”)

u1 v 1 u2

u(x,y)N1(x,y)u1 N2(x,y)u2 N3(x,y)u3N4(x,y)u4

v(x,y)N1(x,y)v1 N2(x,y)v2 N3(x,y)v3N4(x,y)v4

u1 v 1 u 2

(x, y) N1

v (x, y) u

0 N2 0 N3 0 N4 0

N1 0 N2 0 N3 0 N4

TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT

Approximation of the strain in element ‘e’

x2u u 3

x3u 4N (x,y) N(x,y) N(x,y) N (x, y)

u(x,y)

v(x,y)

y 1N(x,y)

N(x, y) 3N (x,y)

y3v 4N (x,y)

u (x,y)

yv(x,y)

xN1(x,y)

u N(x,y)1

x1v ......

u1 v N (x,y) N(x,y) N(x,y) N(x,y) 1

1 2 3 4 0 u2

N1(x,y)x

N2(x,y)x

N3(x,y)x

0N4(x,y)v2

N (x,y)y

N1(x,y

N2(x,y)

N3(x,y)

N3(x,y

N4(x,y)

ε B d

Displacement approximation in terms of shape functions

Strain approximation in terms of strain-displacement

ε B d

Stress approximation DBd

Element stiffness matrix matrix

Summary: For each element

k VeB DBdV

Element nodal load vector

SN T dSe

N X dV

Constant Strain Triangle (CST) : Simplest 2D finite element

3 nodes per element

2 dofs per node (each node can move in x- and y- directions)

Hence 6 dofs per element

The displacement approximation in terms of shape functionsis

N3u3u (x,y) N1u1 N2u2

v(x,y) N1v1 N2v 2 N3v3

v 1 u(x, y )

v (x, y ) 0

3 v 2 u 3

u21 N26 d61

0 N 2 0 N 3 0 u 2

N 1 0 N 20

Formula for the shape functions are a b x cyN1

a2 b2 xc2 y2A

where a b x c yN3

A areaof triangle1 1 y1

a1 x2 y3

a2 x3 y1

a3 x1 y2

b1 y2 y3

b2 y3 y1

b3 y1 y2

det 1 x y

1 x 3y3

c1 x3 x2

c2 x1 x3

c3 x2 x1

The shape functions N1, N2 and N3 are linear functions of x and y

Properties of the shape functions

at node 'i'

at other nodes

At every point in the domain

i iN x x

N i y i y

Geometric interpretation of the shape functions, at anypoint P(x,y) that the shape functions are evaluated

N 2 A 2

Approximation of the strains

v B d y

y xu v

N1(x,y)0

N3(x,y)0

N2 (x,y)

xN (x, y)

N (x,y)x

N(x,y)

B 0x 2 3 y

N(x,y)N (x,y)

N (x,y) N (x, y)1 2

N (x,y) N (x, y)2 3 3 x y x y

1 b3 0

2Ac1 c2

b 0 b2 0 0 c 0 c 0 c

3c3 b3

Inside each element, all components of strain are constant: hence the name Constant Strain Triangle.

Element stresses (constant inside each element).

IMPORTANT NOTE:

The displacement field is continuous across element boundaries

The strains and stresses are NOT continuous across element

boundaries

BT D B dVe

Since B is constant

k BTD BVe

dV BTDB At t=thickness of the element A=surface area of the element

Element stiffness matrix

Element nodal load vector

f NT X dV

NT T dSe

Element nodal load vector due to body forces

V e N X dV tAe

N X dA

b1x Aet

1 aN X dA

fb 2 x Ae

N X2 adA

b 2y tAe

b3 y tAe

N 2 Xb dA

N3 Xa dA

N3 Xb dA

fb3 y Ae

N3dA 3

N3 X a

N3 Xb dA

EXAMPLE:

If Xa=1 and Xb=0

N Xe 1 a tA

Ae N1 Xb

t N X dA tA 0

b 2y Aet

dA tAe

N2 X a

dA tAe

Element nodal load vector due to traction

f NT T dSS

EXAMPLE:

NT tl13

eT S dS

along 13

Element nodal load vector due to traction

Example

NTf tS

23e ST dS

along 23

S2 x23

2 along 23(1)dy

Similarly, compute

Recommendations for use of CST

Use in areas where strain gradients are small

Use in mesh transition areas (fine mesh to coarse mesh)

Avoid CST in critical areas of structures (e.g., stress concentrations,

edges of holes, corners)

In general CSTs are not recommended for general analysis

purposes as a very large number of these elements are required

for reasonable accuracy.

Example

Thickness (t) = 0.5 in E= 30×106 psi n=0.25

(a) Compute the unknown nodal displacements.

(b) Compute the stresses in the two elements.

Nodal coordinates

Realize that this is a plane stress problem and therefore we needto use

E 3.2 0 7

0 0.80

D 1 2 1

3.2 0 10 psi

0 0 1.2 2

Step 1: Node-element connectivity chart

ELEMEN

(sqin)

1 1 2 4 3

2 3 4 2 3

Node x y

For Element #2:2

B(2) 1 0

00 0 0 2

63 0 3 0

3 2 3 0 0 2

Recall

0 b 0 b2 3

1 2 3with

b1 y2 y3

c1 x3 x2

b2 y3 y1

c2 x1 x3

b3 y1 y2

c3 x2 x1

1c b1 b2

B 0 c 0 c 0 c2A c c

2 3 b 3

For Element #1: 2(2) y1 0; y2

x1 3;x2

2; y3 0

3; x3 0

Henceb1 2

4(3) 1(1) 2 0 (local numbers within brackets) Therefore

B(1) 1 0

0 0 0 2

3 0 3 0

2 3 0 0

Step 2: Compute strain-displacement matrices for the elements

Step 3: Compute element stiffness matrices

AtB Dk(1) B(1)

(3)(0.5)B D B(1)

0.9833 0.5 0.3

0.3107

u1 v1 u2 v2 u4

0.2 v4

0.45 0.2 0.5333

1.4 0.3 1.2 0.2

0.45 0 0

1.2 0.2 0

0.5333

k(2) AtB(2)T D B(2)

(3)(0.5)B(2) D B(2)

0.9833 0.5 0.3

1.4 0.3

0.3107

u3 v3u4 v4

0.2 0.5333

1.2 0.2

0.5333

0.2 v2

Step 4: Assemble the global stiffness matrix corresponding to thenonzero degrees of freedom

• u3 v3 u4 v4 v1 0

• Hence we need to calculate only a small (3x3) stiffness matrix

0.983 0.45 0.2 u0.45 0.983

0 1071u

x03 32

2 y fS N (300)tdx

(300)(0.5) 3

x03 32

50 2 50 2 225 lb

f f1x 0

2 y 2 y

The consistent nodal load due to traction on the edge 3-2

1 0 0 0 fS 2 y

Hence f2 y 1000 f

1225 lb

Step 6: Solve the system equations to obtain the unknownnodalloads Kd f

0.2u1 0 107 0.983

0.450.45

0.983 0.2 0 1.4 v 1225

Solve to get

u 0.2337104 in 1

u2 0.1069104 in

2 0.9084104 in

Step 7: Compute the stresses in the elements

In Element #1

(1) DB(1) d(1)

Calculate

u v u1 1 2

0.2337104

v u4 v42

0 0.1069104 0.9084104 0 0

(1) 1391.1

In Element #2

(2) DB(2) d(2)

d(2)T u3 4 2 2v

Calculate

v3 u4 v u

0 0 0 0 0.1069104 0.9084104

28.52 psi

363.35

Notice that the stresses are constant in each element

Axi-symmetric Problems

Definition:

A problem in which geometry, loadings, boundary conditions and materials are symmetric about one axis.

Axi-symmetric Analysis

x rcos; y rsin; z z

• quantities depend on r and z only

• 3-D problem

• 2-D problem

Axi-symmetric Analysis

Axi-symmetric Analysis – Single-Variable Problem

ra u f (r, z) 0

r r 11 r

u(r, z) u(r, z)

Weak form:w u w u wu 0 r

22 a00

11 r z

wf (r, z)rdrdz

wqn dse

q a u(r, z)n a

u(r, z)n

Finite Element Model – Single-Variable Problem

jwhere j (r, z) j (x, y)

Ritz method:

Weak form

Keue f e Qe

ij j i i

ij a11i ja

i j rdrdzr r 22

00 i j

f e frdrdze

i i n q ds

Single-Variable Problem – Heat Transfer

Heat Transfer:

rk T(r, z) T(r, z) f (r, z) 0r r r z k z

Weak form

r r z z0 w k

T wf (r, z)rdrdz

wqndse

where q k T(r, z)n k

T (r, z)n

n rr z

3-Node Axi-symmetric Element

T (r, z) T11T22 T33

1 r zr z r z2 3 3 2

1 z z2A

e r2 r3

1 r zr z rz

3 1 1 3

2A 3 1

1 r zr z r z

2 2 13

2A 1 2 er r

4-Node Axi-symmetric Element

T (r, z) T11T22 T33 T44

2 a b 1 a b

3 a b 4 1

Single-Variable Problem –Example

Step 1: Discretization

Step 2: Element equation

z rdrdz

i i n q dse

Qe f e frdrdz

Review of CST Element

Constant Strain Triangle (CST) - easiest and simplest finite element

Displacement field in terms of generalized coordinates

Resulting strain field is

Strains do not vary within the element. Hence, the name constant strain triangle (CST)•Other elements are not so lucky.•Can also be called linear triangle because displacement field is linear in x and y - sides remain straight.

Constant Strain Triangle

The strain field from the shape functions looks like:

– Where, xi and yi are nodal coordinates (i=1, 2, 3)

– xij = xi - xj and yij=yi - yj

– 2A is twice the area of the triangle, 2A = x21y31-x31y21

Node numbering is arbitrary except that the sequence 123 must go clockwise around the element if A is to be positive.

Constant Strain Triangle

Stiffness matrix for element k =BTEB tA

The CST gives good results in regions of the FE model where

there is little strain gradient

• Otherwise it does not work well.

Linear Strain Triangle

Changes the shape functions and results in quadratic displacement distributions and linear strain distributions within the element.

Example Problem

Consider the problem we were looking at:

I 0.113 / 12 0.008333in4

I 0.008333 60 ksi

0.00207E

2EI 2 29000 0.008333 0.0517in.

UNIT-IVSTEADY STATE HEAT TRANSFER ANALYSIS

CLOS Course Learning Outcomes

CLO 1 Understand the concepts of steady state heat transfer analysis

for one dimensional slab, fin and thin plate.

CLO 2 Derive the stiffness matrix for fin element.

CLO 3 Solve the steady state heat transfer problems for fin and

composite slab.

Thermal Convection

Newton’s Law of Cooling

q h(Ts T)

h: convective heat transfer coefficient (W m2 Co )

Thermal Conduction in 1-D

Boundary conditions:

Dirichlet BC

Natural BC

Mixed BC

Governing Equation of 1-D Heat Conduction

d (x) A(x)

dT (x) AQ( x) 0 0<x<Ldx dx

Weighted Integral FormulationL

d dT (x) 0 w(x) dx

(x) A(x)dx

AQ( x) dx

Weak Form from Integration-by-Parts

L dw dT w

dx wAQ dx A

dx 0 0

Weak Formulation of 1-D Heat Conduction (Steady State Analysis)

Formulation for 1-D Linear Element

T(x) T11(x)T22(x)

1 ( x) 2 ,2 ( x)

x x x x1

f (x) AT

, f (x) AT

Let w(x)= fi (x), i = 1, 2

j dx 2 AQdx (x )fj dx dx

i i 2 2

(x ) f i 1 1

j1 x 1

Qi i (x2 ) f2 i (x1 ) f1

2 Q2 K12

11 K12 T1

K22 T2

where K Aij

i j dx dx

dx, Q AQdx, f A

dx, f A

Element Equations of 1-D Linear Element

A 1 f1 Q1

where Q

AQ dx, f Ai 1

, f2Adx

1-D Heat Conduction - Example

A composite wall consists of three materials, as shown in the figurebelow. The inside wall temperature is 200oC and the outside airtemperature is 50oCwith a convection coefficient of h = 10 W(m2.K).Find the temperature alongthe composite wall.

1 70W m K, 2 40W mK, 3 20W m K

t1 2cm, t22.5cm, t3 4cm

Thermal Conduction and Convection- Fin

Objective: to enhance heat transfer

Governing equation for 1-D heat transfer in thin fin

dT A Q 0c dx

Q 2h(T T) dx w 2h(T T) dx t

2h(T T ) w t loss A dx A

d A dT

Ph T T A Q 0

dx cdx

where P 2 w t

Fin (Steady State Analysis)

Governing Equation of 1-D Heat Conduction

d (x) A(x)

dT (x) Ph T T AQ 0 0<x<Ldx

Weighted Integral FormulationL

d dT (x) 0 w(x) dx

(x) A(x)dx

Ph(T T) AQ( x) dx

Weak Form from Integration-by-PartsL dw dT

0 dx A

dx wPh(T T) wAQ dx w A

xx1 x x2

Let w(x)= fi (x), i = 1, 2

T x2 A di d j Ph dx x2 AQ PhT dx

dx dx i

j1j x i j

i (x2 ) f2 i (x1) f1

Qi i (x2) f2 i (x1) f1

K11 K12 T1

2 Q2 K12

K22 T2

where K x2 A di d j Ph dx, Q x2 AQ PhT dx,ij dx dx i j i i

, f2 A

f1 Q1 A 1 1 1 T1Phl 2

2 Q2 L 1 1 6 1 2T2

Element Equations of 1-D Linear Element

Qi i AQ PhTdx,x1

f1 Adx

, f2 Adx

In general ux, t

Two approaches:

ux,tu j j x, t

ux,tu j tj x

Time-Dependent Problems

1 1 2 2

Weak form:

w u0 a x x t cw

dx Q w(x )Q w(x )

Model Problem I – Transient Heat Conduction

f x, t

1 1 2 2

let: ux,tu j t j xj 1

and iw x

wu0 a x x tcw

dx Q w(x ) Q w(x )

K uM uFx2

K a i j dx Mij

ci jdx

Fi i fdx Qi

Transient Heat Conduction

Time Approximation – First Order ODE

k1 kabt

dt bu ft 0 t T u0 u0

Forward difference approximation – explicit

ak k k

u t f bu

Backward difference approximation - implicit

u u t f bu

Stability Requirement

2t tcri 1 2 max

where K M uQ

Note: One must use the same discretization for solvingthe eigenvalue problem.

Transient Heat Conduction - Example

t x2 0 0 x 1

u 1,t 0

ux,01.0

UNIT-VDYNAMIC ANALYSIS

DYNAMIC EQUATIONS

P = ku K = fP

Stiffness and flexibility stiffeness matrix

Dynamic equations

Vibration analysis

The Vibration Eigen problem

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